题目：
Given a binary tree, return the level order traversal of its nodes' values. (ie, from left to right, level by level).
For example:
Given binary tree {3,9,20,#,#,15,7},
3
/
9 20
/
15 7
return its level order traversal as:
[ [3], [9,20], [15,7] ]
解题思路：
广度优先遍历。
代码：
1 /** 2 * Definition for binary tree 3 * struct TreeNode { 4 * int val; 5 * TreeNode *left; 6 * TreeNode *right; 7 * TreeNode(int x) : val(x), left(NULL), right(NULL) {} 8 * }; 9 */ 10 class Solution { 11 public: 12 vector<vector<int> > levelOrder(TreeNode *root) { 13 vector<vector<int>> ans; 14 if (root == NULL) return ans; 15 16 queue<TreeNode *> one, another; 17 one.push(root); 18 19 while (!one.empty() || !another.empty()) { 20 if (!one.empty()) { 21 vector<int> cur_ans; 22 while (!one.empty()) { 23 TreeNode * node = one.front(); 24 cur_ans.push_back(node->val); 25 if (node->left != NULL) another.push(node->left); 26 if (node->right != NULL) another.push(node->right); 27 one.pop(); 28 } 29 ans.push_back(cur_ans); 30 } 31 if (!another.empty()) { 32 vector<int> cur_ans; 33 while (!another.empty()) { 34 TreeNode *node = another.front(); 35 cur_ans.push_back(node->val); 36 if (node->left != NULL) one.push(node->left); 37 if (node->right != NULL) one.push(node->right); 38 another.pop(); 39 } 40 ans.push_back(cur_ans); 41 } 42 } 43 return ans; 44 } 45 };